Question

Find the mean, mode and median.

Marks	No. of students
\[25 - 35\]	\[7\]
\[35 - 45\]	\[31\]
\[45 - 55\]	\[33\]
\[55 - 65\]	\[17\]
\[65 - 75\]	\[11\]
\[75 - 85\]	\[1\]

Answer 1

Hint: Use the formulas of Mean, Median for the grouped data, also use \[Mode = 3(media) - 2(mean)\] for evaluation of Mode.

Complete step-by-step answer:
The mean of observations as we know is the sum of the values of all the observations divided by the total number of observations.
\[Mean\,\left( {\overline X } \right) = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\]
Here Greek letter \[\sum {} \] means summations and \[{f_i}\] and \[{x_i}\] are the respective class mark and frequency for each class interval we require a point which serves as the representative of the whole class. It is assumed that the frequency of each class is centered on its mid-point.
So the midpoint of each class can be chosen to represent the observations failing in the class.
Recall that we find the midpoint of a class (or its class mark) by finding the average of its upper and lower limits.
That is \[Class\,\,mark\,\, = \,\,\dfrac{{Upper\,\,Class\,\,Limit\,\, + \,\,Lower\,\,Class\,\,Limit}}{2}\]
With reference to the table the class mark for internal \[25 - 35\] is \[\dfrac{{25 + 35}}{2} = 30\]. So let's prepare the whole table as shown:

Marks	No. Of Students(fi)	Class Marks(xi)	(fixi)
\[25 - 35\]	\[7\]	\[30\]	\[210\]
\[35 - 45\]	\[31\]	\[40\]	\[1240\]
\[45 - 55\]	\[33\]	\[50\]	\[1650\]
\[55 - 65\]	\[17\]	\[60\]	\[1020\]
\[65 - 75\]	\[11\]	\[70\]	770
\[75 - 85\]	\[1\]	\[80\]	\[80\]

Adding the number of students and the values of \[{f_i}{x_i}\], we’ll get:
\[
   \Rightarrow \sum {{f_i} = 100} \\
   \Rightarrow \sum {{f_i}{x_i} = 4979} \\
 \]
\[{f_i}{x_i}\] is obtained by Multiplying frequency column with class mark column.
Thus we have:
\[
   \Rightarrow Mean\,\left( {\overline X } \right) = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} \\
   \Rightarrow Mean\,\left( {\overline X } \right) = \dfrac{{4970}}{{100}} \\
   \Rightarrow Mean\,\left( {\overline X } \right) = 49.7 \\
 \]
Now for calculation of median, we require cumulative frequency (cf) which is nothing but the sum of frequencies preceding it. Let us prepare the table first.

Marks	No. Of Students(fi)	Cumulative Frequency(cf)
\[25 - 35\]	\[7\]	\[7\]
\[35 - 45\]	\[31\]	\[38\]
\[45 - 55\]	\[33\]	\[71\]
\[55 - 65\]	\[17\]	\[88\]
\[65 - 75\]	\[11\]	\[89\]
\[75 - 85\]	\[1\]	\[100\]

So we again have:
\[
   \Rightarrow N = \sum {{f_i} = 100} \\
   \Rightarrow \dfrac{N}{2} = 50 \\
 \]
From the table, the cumulative frequency just greater than \[50\] is \[71\] and the corresponding class is \[45 - 55\] which will become our median class.
According to the above table, we have:
\[
   \Rightarrow \,l\,\, = \,\,45 \\
   \Rightarrow h\,\, = \,\,10 \\
   \Rightarrow N\,\, = \,\,100 \\
   \Rightarrow f\,\, = \,\,33 \\
   \Rightarrow cf\,\, = \,\,38 \\
 \]
Now, median can be calculated using the formula as shown:
\[ \Rightarrow median\,\, = \,\,l\,\, + \,\left( {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right) \times h\]
Putting all values, we’ll get:
\[
   \Rightarrow median = 45 + \left( {\dfrac{{50 - 38}}{{33}}} \right) \times 10 \\
   \Rightarrow median = 45 + 3.64 \\
   \Rightarrow median = 48.64 \\
 \]
Now using \[Mode = 3(media) - 2(mean)\], well get
\[
   \Rightarrow Mode = 3 \times 48.64 - 2 \times 49.70 \\
   \Rightarrow Mode = 145.92 - 99.4 \\
   \Rightarrow Mode = 46.52 \\
 \]
So the value of mean is 49.7, the value of median is 48.64 and the value of mode is 46.52

Note: For converting ungrouped data into grouped data, the first step is to determine how many classes you want to have. Next, you subtract the lowest value in the data set from the highest value in the data set and then you divide by the number of classes that you want to have.